sin(n) is dense!

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[xizarrg]

"Super duper close" is an excellent explainer for "dense".

[TheEternalVortex]

We can generalize Lemma 2 to note that { a + b y | a, b ∈ ℤ and y is irrational } is dense. And likewise any continuous function on a dense set is dense on its codomain. So then you could use that to perhaps find other interesting examples of dense functions on integers.

We can also generalize the claim made as part of the proof to say that if G (a subgroup of R) has a least positive element then it's cyclic, which is kind of interesting by itself.

[tiripoulain]

It’s not immediately clear that there is 0<g<ε in G. Let’s say that G is not trivial.

Let G_+ denote the strictly positive part of G. Let m=inf G_+. By assumption m is not in G_+, and m>=0 because 0 is a lower bound.

There is a sequence (g_n) of points in G_+ approaching m, because m is a limit point of G_+. In particular, (g_n) is Cauchy.

So for every ε>0, g:=|g_n-g_m|<ε for n, m large. Because G is an additive subgroup of R, g is an element of G. It is positive, so g is in G_+. (edit: g can be taken to be >0 because the sequence is not eventually constant -- otherwise m in G, contradicting the assumption)

Therefore, for every ε>0, there is g in G_+ such that:

0 <= m < g < ε.

Thus m=0, and G_+ has arbitrarily small elements.

[albertogarcia]

In 10:58 is asssuming that nd≠0. It needs some justification. Notice if nd=0 then 1=md, with m, d integers, so the unique way it happens is when m, d are units in Z, i.e. m=±1 and d=±1, but then g=m (cause dg=md) and G=<g>=<±1>=Z, which is not true since G contains irrational numbers like 2π when a=0 and b=1.

[Bodyknock]

2:14 “Since G does not have a least positive element we should be able to find a g in G such that g < E for a given E.”

That actually sounds like it needs to be proven. Hypothetically, for example, not knowing anything else about groups or G, it may be that the elements of G form a set that has an infinimum but not a minimum (e.g. looks like numbers of the form 1 + 1/2^n plus 0. That set, if it were a group, would have no least positive element (assuming we’re not defining 0 as positive) but its infimum would be 1, not 0, so of E=1/4 for example there would be no positive element of G less than E.

Note that I’m not saying the claim in the video is wrong, I’m just saying it doesn’t immediately follow at face value from the definitions in the video, it’s something that should be proven.

[lucid_]

First time seeing an XOR lemma in maths, so cool!

[Pauliwallnutz]

This is the most confusing thing I have ever witnessed, having never taken math courses beyond linear algebra, d.e., and calc 3

[zetacrucis]

Sweet. I went to a maths specialist high school and one of the coolest things I remember from the year 9 syllabus was properties of sin(n) for positive integers n: things like how it's always irrational, pseudorandom, never repeats (one-to-one), and indeed that it's dense on [-1:1].

[AdamGlesser]

A plug for the hole in the first argument in Lemma 1. Assume that (0, ε) and G are disjoint. Since G is a group, the difference of any two elements of G is an element of G, and hence |g-h| >= ε for all g, h in G.

Let N be the minimal positive integer such that there exists positive g in G such that g < Nε. This exists by the axiom of Archimedes as long as we assume that G contains at least one positive element.

As G contains no least positive element, there exists h in G such that 0 < h < g. Thus, g-h >= ε or, equivalently, h+ε <= g < Nε, and hence h < (N-1)ε. This contradicts the minimality of N. Thus, G intersects (0, ε).

[yurenchu]

11:00 I think it needs to be pointed out explicitly that g must be smaller than 1 (by showing that G contains at least one element that is between 0 and 1) _and therefore_ n cannot be 0 .

Because if n can be 0, then it may be possible to write 1 as d*g without violating the irrationality of π .

[ivanklimov]

one of the best videos on this channel yet, and that's saying a lot. especially nice how the density of G in the second lemma boiled down to pi being irrational

[benhsu]

Thanks! Really liked this video and your other recent one about Fermat Little Theorem

[zVepox]

There might be another solution:
Consider m, n integers and x irrational.
Further consider k as a natural number and the list of numbers <<0>>, <<x>>, …, <<kx>> and note they all are different, as x is not rational. „<<…>>“ denotes the fractional part.
We then split the interval [0, 1) in k parts of equal length, so there are two numbers in our list in one section. Hence
0< <<rx>>-<<sx>> <1/k => 0< <<(r-s)x>> < 1/k => 0<m+nx<1/k, where m = r-s and n=floor(mx).

Hence for all intervals (a, b), y:=m+nx is contained in (0, b-a) for some m, n.
Therefore, (b-a)/y > 1 and (a/y, b/y) has a length greater than 1, so it must contain an integer l.
=>lm+lnx is contained in (a, b).

Write sin(n) = sin(n+2pim) = sin(x), where x can be in any interval for suiting integers m and n.
*because pi is irrational
Hence sin(n) takes all values sin(x) takes eventually, which concludes the proof.

If there us anything wrong with this, just let me know

[StratosFair]

Very nice explanation of a beautiful result !

[ronald]

An alternative proof:

First use the continued fraction expansion of 2π to construct rational numbers p_i/q_i that converge to 2π. Now |p_i - 2πq_i| converges to 0 (because a theorem about continued fractions says so; it does not already follow from p_i/q_i converging to 2π).

To approximate x = sin(theta) in [-1, 1], choose integers a_i such that a_i(p_i-2πq_i) is closest to theta. Note that the difference with theta will be smaller than |p_i-2πq_i|. Now take n_i = a_i * p_i, and we will have sin(n_i) = sin(a_i(p_i-2πq_i)) -> sin(theta) = x.

Since for any x in [-1, 1] we can find a sequence n_i with sin(n_i) -> x, it follows that sin(N) is dense in [-1, 1].

[aleksandervadla]

Think you also can use Weil’s Equidistribution theorem

[geryz]

10:39 Why not just set md = 1 and nd = 0 ? Possible if you set d = 1, m = 1, n = 0
Then also pi = 0/0 which isn't a rational number anymore

[VinyJones]

11:15 would have be nice to prouve n≠0, it is by contradiction because then g=1 but 0 < 7-2π < 1=g so g is not the least positive element of g

[daliasprints]

Interesting aside: a numerical floating point version of it isn't if you don't do proper argument reduction.

[kumoyuki]

that's quite a nice proof. the proposition seems obvious, but I'm sure that I would have had a hard time formalizing it. Thank you :)